Question

1. Prove that no proper subset of RRR is simultaneously open and closed in the RR topology). 2. Prove that no proper subset of RFC is simultaneously open and closed.

Answer 1

Solution Given data, here, he have to prove that no proper Subset of Rer is Simultaneously open and closed So, RR- {UER} V TEV, JA ER Such that selatos) som qe AG RRR which is open subset and closed A=U (artoo) GEA C Now at a ine {at A} As (a*, too) Similarly pop and is also closed AC is open But AC (-00, ax) is open then tao 6 -ỳ (a* Độc BC contradiction to Cat; oo) = A It is AC

Therefore, finally ma is either we can open loro say that closed best not both at a time. (2) Here, we have to prove that no proper subset of Rec is simultaneously open and closed. Now, as Rfc & A is closed and open qpe "A" is open, Then A: IRIQ, ---- an} & B:fa, ---- and SR AC is open and Bi{a,... an} is open closed, then But "B" is not open in Rfc We know BER 1261,---- bn} for any subset of ļ bogacocobon}

437. 7 "B" is fenate then "B" is not open 98 "A" is not closed & open at Hvoe Therefore, Pfc has no proper subset which is open and closed simentlaneously. Hence Solved